Problem: A box contains $3$ red balls, $4$ green balls, and $3$ blue balls. If we choose a ball, then another ball without putting the first one back in the box, what is the probability that the first ball will be red and the second will be green? Write your answer as a simplified fraction.
Solution: The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened In this case, event A is picking a red ball and leaving it out. Event B is picking a green ball. Let's take the events one at at time. What is the probability that the first ball chosen will be red? There are $3$ red balls, and $10$ total, so the probability we will pick a red ball is $\dfrac{3} {10}$ After we take out the first ball, we don't put it back in, so there are only $9$ balls left. Since the first ball was red, there are still $4$ green balls left. So, the probability of picking a green ball after taking out a red ball is $\dfrac{4} {9}$ Therefore, the probability of picking a red ball, then a green ball is $\dfrac{3}{10} \cdot \dfrac{4}{9} = \dfrac{2}{15}$